Question

Tìm x biết

a) \(\sqrt{x^2-3}\le x^2-3\)

b) \(\sqrt{x-1}< x+3\)

c) \(\sqrt{x^2-6x+9}>x-6\)

Rút gọn: \(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\) ; 0 \(\le x\le y\)

Answer 1

b: \(\sqrt{x-1}< x+3\)

nên \(\left\{{}\begin{matrix}x-1>=0\\\left(x-1\right)^2< \left(x+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-2x+1-x^2-6x-9< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\-8x-8< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\-8x< 8\end{matrix}\right.\Leftrightarrow x>=1\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=6\\x^2-6x+9>x^2-12x+36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=6\\6x>27\end{matrix}\right.\Leftrightarrow x>=6\)

Bài 2: 

\(=\sqrt{\left(x-y\right)^2}=\left|x-y\right|=y-x\)